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I have this schematics here enter image description here

Does anyone know how the capacitor in the black circle (disregard the blue circles) can discharge once it is charged up and the power has been disconnected.

It is suggested that it is discharged through the resistor but this cant be because once it is charged its left plate will be +5Vcc and the right plate 0V so to discharge through the resistor the charge would have to go through the insulation through the resistor and to ground?? Witch is impossible unless the capacitor is bad.

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  • \$\begingroup\$ There is NO ideal capacitor. The real equivalent is with a resistor (high value) in parallel with capacitance, which is the insulator between plates resistance. So there is leakage current all the time. Typically in in electrolytic capcitors the leakage current is around 5~15μA per μF, which is really high! \$\endgroup\$
    – GR Tech
    Feb 2, 2015 at 8:51
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    \$\begingroup\$ Thinking and questioning are both extremely good ideas - they will take you far. BUT do also listen to what people say and work through stuff you do not understand. If in doubt ask. Ambiorix's "-5V" explanation is essentially correct but you tend to reject it as unlikely, without having any reason to do so than intuition. In this case intuition fails. The voltage across a capacitor cannot change instantaneously nor if no current flows. At power-off there is +5V across the cap. When V+ falls to 0V there is still 5V across the cap so the bottom must be at -5V. In practice things like IC body ... \$\endgroup\$
    – Russell McMahon
    Feb 2, 2015 at 9:34
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    \$\begingroup\$ ... diodes and other muddy the waters but that shos the basic principle. \$\endgroup\$
    – Russell McMahon
    Feb 2, 2015 at 9:35
  • \$\begingroup\$ i did not reject -5v explanation, look below :) \$\endgroup\$
    – adoion
    Feb 2, 2015 at 10:36

1 Answer 1

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It will discharge when the +Vcc is switched off. The left plate is then connected to the ground via the power supply. The RST input then becomes -Vcc with respect to the ground and discharges via the resistor. +Vcc present:

enter image description here

+Vcc not present:

enter image description here

When the capacitor is charged the left plate is +5V with respect to the right plate. Since the +5Vcc power supply is connected to the left plate of the capacitor with the opposite polarity, the RST pin is +5v-(+5V) = 0V. However when you cut the power supply the left plate is connected to the GND (0V). RST then becomes 0V - (+5V) = -5V.

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  • \$\begingroup\$ Why would the RST pin become -Vcc when it is 0V after capacitor charge up? \$\endgroup\$
    – adoion
    Feb 2, 2015 at 8:14
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    \$\begingroup\$ And take the simplest Voltage regulator possible a resistor in series with a Zener diode, if the input is cut of then the capacitor on the output would still reverse bias the Zener diode and would discharge down to the breakdown voltage of the diode but not fully. \$\endgroup\$
    – adoion
    Feb 2, 2015 at 8:25
  • \$\begingroup\$ Even if the PS has no leakage resistance at all, the other components in the circuit do. \$\endgroup\$
    – Ambiorix
    Feb 2, 2015 at 8:42
  • \$\begingroup\$ Ok, I think I made some sense of it, when the power is cut of and if the capacitor is then connected via power supply to ground then the left plate will loose charge while the right plate will pull charge from the ground through the resistor, right ? \$\endgroup\$
    – adoion
    Feb 2, 2015 at 8:43
  • \$\begingroup\$ That's correct. \$\endgroup\$
    – Ambiorix
    Feb 2, 2015 at 9:04

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